91Ó°ÊÓ

Integration is a powerful mathematical tool utilized to reverse the process of differentiation. In the context of first-order linear differential equations, integration plays a central role. In our problem, we are given an equation of the form \( \frac{d}{dt}(a(t)y(t)) = f(t) \), which implies that we can find a solution through integration.

When both sides of this differential equation are integrated with respect to \( t \), we essentially capture the cumulative sum of small differences, thus obtaining a relationship for \( a(t)y(t) \). This allows us to express \( y(t) \) in terms of an integral involving known functions. Consider the integral:

• On the left-hand side: \( \int \frac{d}{dt} (a(t)y(t)) \, dt \) — which simplifies directly to \( a(t)y(t) \) as integration reverses differentiation.
• On the right-hand side: \( \int f(t) \, dt \) — which requires evaluation to find a specific expression in terms of \( t \).

This method helps in determining the general solution of the differential equation by expressing \(y(t)\) in a more solvable form. Integration, therefore, acts like a bridge that transforms our differential equation into a solvable algebraic form.

An initial value problem is a differential equation accompanied by an initial condition, which specifies the value of the solution at a particular point. It is a common scenario in mathematical modeling where we need both a general solution and a particular solution that fits specific practical problems.

In this exercise, we started with the equation involving \( y'(t) \) and given the initial condition \( y(1) = 6 \). The process of solving involves, first, finding the general solution of the differential equation. Next, we apply the initial condition to these general forms to determine the arbitrary constant.

• Determine initial conditions: Use given parameters like \( y(1) = 6 \) to find unique constants.
• Find constants: In general solutions, substitute these initial conditions to solve for constants. For instance, setting all terms to when \( t = 1 \), we were able to solve for the constant \( C \).

By deploying initial conditions, the solution to a differential equation becomes specific and applicable to real-world scenarios where exact initial states are known.

The derivative of a product is an essential concept in calculus that helps deal with functions written as products of two or more simpler functions. The product rule for differentiation states that for two functions \( u(t) \) and \( v(t) \), the derivative with respect to \( t \) is: \( \frac{d}{dt}(u(t)v(t)) = u'(t)v(t) + u(t)v'(t) \).

In applying this to the form of our differential equation \( a(t)y'(t) + a'(t)y(t) \), we recognize that we are dealing with a derivative of the product \( a(t)y(t) \).

• Identify a product form: Recognize expressions like \( a(t)y(t) \) that fit the product rule.
• Apply derivative as a product: Notice the expression resembles \( u'(t)v(t) + u(t)v'(t) \), implying \( \frac{d}{dt}(a(t)y(t)) \).

Understanding derivatives of a product helps to simplify complex differential equations to a more solvable form. It is a crucial step in turning our differential equation into the integrable form \( \frac{d}{dt}(a(t)y(t)) = f(t) \), easing the integration process.